Let r(x)=f(g(h(x))), where h(1)=2, g(2)=3, h'(1)=4, g'(2)=5, and f'(3)=6, how do you find r'(1)?

Question

Christopher Allers · Accepted Answer

The value of #r'(1)# is #120#.

The chain rule for two functions is #d/dx(f(g(x)) = f'(g(x)) * g'(x)# So it would make sense if #d/dx(f(g(h(x))) = h'(x) * g'(h(x)) * f'(g(h(x))# We don't have any concrete functions here to work with, but we do know the values of the functions/derivatives at certain points. #r'(x) = h'(x) * g'(h(x)) * f'(g(h(x))# #r'(1) = h'(1) * g'(h(1)) * f'(g(h(1)))# #r'(1) = 4 * g'(2) * f'(g(2))# #r'(1) = 4 * 5 * f'(3)# #r'(1) = 4 * 5 * 6# #r'(1) = 120# Hopefully this helps!

Evelyn Agard · Answer

undefined To find $ r'(1) $, use the chain rule:

$$ r'(x) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) $$

Evaluate each derivative at $ x = 1 $:

$$ h(1) = 2, \quad g(2) = 3, \quad h'(1) = 4, \quad g'(2) = 5, \quad f'(3) = 6 $$

$$ r'(1) = f'(g(h(1))) \cdot g'(h(1)) \cdot h'(1) $$
$$ = f'(g(2)) \cdot g'(h(1)) \cdot h'(1) $$
$$ = f'(3) \cdot g'(2) \cdot h'(1) $$
$$ = 6 \cdot 5 \cdot 4 $$
$$ = \boxed{120} $$

Paisley Johnson · Answer

undefined To find $ r'(1) $, we can use the chain rule.

Given:
- $ r(x) = f(g(h(x))) $
- $ h(1) = 2 $
- $ g(2) = 3 $
- $ h'(1) = 4 $
- $ g'(2) = 5 $
- $ f'(3) = 6 $

First, we find $ h'(x) $, $ g'(x) $, and $ f'(x) $ using the given derivatives.

Now, apply the chain rule:
$$ r'(x) = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x) $$

Plug in the given values:
$$ r'(1) = f'(g(h(1))) \cdot g'(h(1)) \cdot h'(1) $$

Substitute the given values:
$$ r'(1) = f'(g(2)) \cdot g'(2) \cdot h'(1) $$

We're given $ g(2) = 3 $, so:
$$ r'(1) = f'(3) \cdot g'(2) \cdot h'(1) $$

Substitute the given value for $ f'(3) = 6 $:
$$ r'(1) = 6 \cdot 5 \cdot 4 $$

Calculate the result:
$$ r'(1) = 120 $$

Therefore, $ r'(1) = 120 $.